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According to Descartes' Rule of Signs, can the polynomial function have exactly $5$ positive real zeros, including any repeated zeros? Choose your answer based on the rule only. $\newline$ $f(x) = x^5 - 8x^4 + 8x^3 - 4x^2 + 7$ $\newline$ Choices: $\newline$ (A)yes $\newline$ (B)no

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Descartes' Rule of Signs

Full solution

Q. According to Descartes' Rule of Signs, can the polynomial function have exactly

5

positive real zeros, including any repeated zeros? Choose your answer based on the rule only.

\newline

f(x) = x^5 - 8x^4 + 8x^3 - 4x^2 + 7

\newline

Choices:

\newline

(A)yes

\newline

(B)no

Apply Descartes' Rule of Signs: First, let's apply Descartes' Rule of Signs to $f(x) = x^5 - 8x^4 + 8x^3 - 4x^2 + 7$ .
Count Sign Changes: Count the number of sign changes in the coefficients: $1$ (from $x^5$ to $-8x^4$ ), $2$ (from $-8x^4$ to $8x^3$ ), $3$ (from $8x^3$ to $-4x^2$ ), and $4$ (from $-4x^2$ to $x^5$ $1$ ).
Analyze Possible Real Zeros: According to Descartes' Rule of Signs, the polynomial can have at most $4$ positive real zeros or fewer by an even number.
Determine Real Zeros: So, the polynomial can have $4$ , $2$ , or $0$ positive real zeros.
Conclusion: Since $5$ is not an option, the polynomial cannot have exactly $5$ positive real zeros.

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